时间:2021-07-01 10:21:17 帮助过:13人阅读
MongoDB语法与现有关系型数据库SQL语法比较
MongoDB语法 MySql语法
db.test.find({‘name‘:‘foobar‘})<==> select * from test where name=‘foobar‘
db.test.find() <==> select *from test
db.test.find({‘ID‘:10}).count()<==> select count(*) from test where ID=10
db.test.find().skip(10).limit(20)<==> select * from test limit 10,20
db.test.find({‘ID‘:{$in:[25,35,45]}})<==> select * from test where ID in (25,35,45)
db.test.find().sort({‘ID‘:-1}) <==> select * from test order by IDdesc
db.test.distinct(‘name‘,{‘ID‘:{$lt:20}}) <==> select distinct(name) from testwhere ID<20
db.test.group({key:{‘name‘:true},cond:{‘name‘:‘foo‘},reduce:function(obj,prev){prev.msum+=obj.marks;},initial:{msum:0}}) <==> select name,sum(marks) from testgroup by name
db.test.find(‘this.ID<20‘,{name:1}) <==> select name from test whereID<20
db.test.insert({‘name‘:‘foobar‘,‘age‘:25})<==>insertinto test (‘name‘,‘age‘) values(‘foobar‘,25)
db.test.remove({}) <==> delete * from test
db.test.remove({‘age‘:20}) <==> delete test where age=20
db.test.remove({‘age‘:{$lt:20}}) <==> elete test where age<20
db.test.remove({‘age‘:{$lte:20}}) <==> delete test where age<=20
db.test.remove({‘age‘:{$gt:20}}) <==> delete test where age>20
db.test.remove({‘age‘:{$gte:20}})<==> delete test where age>=20
db.test.remove({‘age‘:{$ne:20}}) <==> delete test where age!=20
db.test.update({‘name‘:‘foobar‘},{$set:{‘age‘:36}})<==> update test set age=36 where name=‘foobar‘
db.test.update({‘name‘:‘foobar‘},{$inc:{‘age‘:3}})<==> update test set age=age+3 where name=‘foobar‘
MongoDB语法
标签:des ini lte order rev 有关 语法 find init